What was missing from your answer was another range of possible values for x, between -3 and + 3

Look at my graph and you can see that, in this range, 'y' is always negative, so certainly < 1/2

So why did these values 'go missing' ?

The graph has a discontinuity at -3 and again at +3 because of the division by zero.

So you need to check in this range too.

I tried to demonstrate why (-3,0) is another set of values that fit the inequality. Twice I had to use the rule that 'if you multiply or divide by a negative, you must reverse the inequality'. Do you know and understand this rule?

I have recently posted an explanation at

[url]http://www.mathisfunforum.com/viewtopic.php?id=18411[/math]

You might like to have a look at that, especially post 7 in that thread.

I've got to log off for a short while. I'll be back on in about 15 mins.

ADDITIONAL EDIT:

case 3 which you actually explained in post #5 gives

but as already given

If I call

... statement one and ... statement two, then we can work the logic like this.

I got statement one by algebra assuming statement two. So statement one is only true to the extent that it obeys statement two. As statement two is a subset of statement one that means the inequality is satisfied just for the subset ie

So that provides part of the missing answer.

Testing x = 0 by substitution shows it may be added to the set giving

You can finish by considering (0,+3) and showing it is legitimate to add this to the rangeBob

Children are not defined by school ...........The FonzYou cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Re: Linear inequation involving absolute value problem...

'if you multiply or divide by a negative, you must reverse the inequality'. Do you know and understand this rule?

Yes,I understand this rule.

You can finish by considering (0,+3) and showing it is legitimate to add this to the range

Well now its a bit more clear,I have to consider/assume the ranges (-3,0) and (0,+3) but I don't understand the logic just because +/-3 makes the denominator 0,in many inequation it happens but we don't have to choose those cases,what my problem was I havent looked at different cases,but still this problem is not clear properly the logic behind choosing the cases ,when I reduced the equation to

from here why I can't able to solve the inequation ?, may be I need more time .

bob bundy what you did I understood totally and appreciate it.Well thanks for helping me out